Trying to put some theory into practice before an exam without any lectured examples, so forgive me if this is simple... I have the subgroup of $GL_2(\mathbb{R})$ given by \begin{equation} G=\bigg{\{}\begin{pmatrix}a & b \\ 0 & 1 \end{pmatrix}:a,b\in\mathbb{R},a\not=0\bigg{\}} \end{equation} which I would firstly like to show is a Lie group under matrix multiplication, a secondly find all its one-parameter subgroups.
It's obvious to me this is not a closed subgroup (so we can't apply Cartan's theorem and conclude $G$ is an embedded Lie subgroup and hence a Lie group), and unlike the subgroups $O_n(\mathbb{R})$ and $SL_n(\mathbb{R})$ etc. I can't write $G$ as the inverse image of a regular value of a smooth map and apply the IFT. How else could I show $G$ is a smooth manifold?
I'm also trying to find all one-parameter subgroups of $G$, i.e. all Lie group homomorphisms $\sigma:\mathbb{R}\rightarrow G$. I've tried writing such a map explicitly in terms of functions $a(t)$ and $b(t)$ and arrive at the conditions $a(t+s)=a(t)a(s)$ and $b(t+s)=a(t)(1+b(s))$, but this doesn't seem very neat. I also have the fact that $G$ is isomorphic as a group to invertible linear transformations $T$ on $\mathbb{R}$ of the form $T(x)=ax+b$ but I don't know if this helps.
Could someone please help me with this example, since all I have at the moment is theory and am struggling to put it into practice. Thanks
As mentioned in the comments, $G$ is a closed subset of $GL_2(\Bbb R)$.
The one-parameter subgroups are exactly the maps $\Bbb R \to G$ defined by $$\phi_X : t \mapsto \exp(t X),$$ for $X \in \mathfrak g \cong T_I G$.
In our case, it's straightforward to check that under this identification (and the identification of $T_I G$ as a subspace of $T_I GL_2(\Bbb R) = T_I M(2, \Bbb R) \cong M(2, \Bbb R)$) the Lie algebra $\mathfrak g$ of $G$ $$\mathfrak g = \left\{\pmatrix{a&b\\0&0} : a, b \in \Bbb R \right\},$$ so it remains just to produce an explicit formula for $$\exp\left[ t \pmatrix{a&b\\0&0\\}\right] = \exp\pmatrix{at&bt\\0&0} .$$ One way to handle this is to write the matrix on the r.h.s. as $A + B$, where $$A := \pmatrix{at&0\\0&0}, \qquad B := \pmatrix{0&bt\\0&0} ,$$ which enables simplification of the expression for the exponential when using the formula $\exp X = 1 + X + \frac{1}{2} X^2 + \frac{1}{6} X^3 + \cdots$ that holds in the matrix group setting.